We define an operation a∗b=a+b−2a * b = a+b-2 over R2\mathbb R^2 for every a,b∈Ra,b\in \mathbb R.

I have to prove that there is no identity element for the operation.

While my intuition and my efforts show time after time that the identity element is 2, I would appreciate an external view of the matter.

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What leads you to believe that there is no identity element?

– Sloan

Oct 20 at 19:19

1

Suppose a+x−2=x+a−2=aa + x -2 = x + a – 2 = a then x=2x =2 … and that’s it. That’s the identity. But I strongly suspect that isn’t the actual question. In particular “on top of R^2” sounds like you have on operation on ordered pairs (a,b) ∈R2\in \mathbb R^2. But … then that isn’t a binary operation and it doesn’t map to R2R^2. What exactly is the question as stated?

– fleablood

Oct 20 at 19:22

In that context the question doesn’t make any sense. “f:X→Yf:X \rightarrow Y. X=R2X = \mathbb R^2 and Y=RY = \mathbb R. f(a,b)=a+b−2f(a,b) = a+b -2. Prove ff has no identity element.” What does mean? It doesn’t make any sense.

– fleablood

Oct 20 at 19:38

the operation is defined for R2, but a and b are members of R, as Omnomnomnom descirbed it.

– Hector Isidor

Oct 20 at 19:44

In case the op is in R, then the approach is clear and simple, while with the team help, we managed to describe the question properly, how to we approach it ?

– Hector Isidor

Oct 20 at 19:51

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3 Answers

3

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If a,b∈Ra,b \in \mathbb R and a∗b=a+b−2a*b = a+ b – 2 then for all a∈Ra \in \mathbb R if a∗x=x∗a=aa*x = x*a =a then a+x−2=x+a−2=aa+x -2 = x +a -2 =a and x=2x =2. That’s true for all a∈Ra \in \mathbb R and it is the only value where it is true so 22 is the identity element.

BUT if a,b∈R2a,b \in \mathbb R^2 and a∗b=a+b−2=(a1,a2)+(b1,b2)+(2,0)=(a1+b1+2,a2+b2)a*b = a+ b – 2= (a_1,a_2) + (b_1,b_2) + (2,0) = (a_1+b_1 + 2,a_2 + b_2) and a+x−2=aa + x – 2=a then (a1+x1+2,a2+z2)=(a1,a2)(a_1 + x_1 + 2, a_2+z_2) = (a_1,a_2) so (x1,x2)=(2,0)(x_1, x_2) = (2,0) and … I got nothing.

But if this is a unary operation…. (a,b)∈R2(a,b) \in \mathbb R^2 and f(a,b)=a+b−2∈Rf(a,b) = a+b -2 \in \mathbb R …. but then it’s a) unary not binary so identities don’t make sense as we don’t every operate on two elements for one to “invert” the other and b) it’s not an “operation” as it doesn’t map to the original set R2\mathbb R^2. So the question makes no sense in that terms.

So, I got nothing. I think the question must mean something else. I’m particularly concerned about that near non-sense phrase “on top of R2\mathbb R^2”. I don’t know what that is supposed to mean in this context.

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Okay, the only real way I see this question as making sense is not as an algebra question at all and not about the identity or inverse elements of a group but as a prove that as a function from f:R2→Rf:\mathbb R^2 \rightarrow R there is no inverse function f−1:R→R2f^{-1}:\mathbb R \rightarrow R^2 so that f−1(f(a,b))=(a,b)f^{-1}(f(a,b)) = (a,b).

For any x∈Rx \in R there are multiple (a,b)(a,b) so that f(a,b)=xf(a,b) =x so there can’t be any f−1f^{-1} that magically “knows” that f−1(x)f^{-1}(x) is supposed to equal (a,b)(a,b) as opposed to (a+e,b−e)(a+e, b-e). i.e. For any x∈Rx \in R there are infinitely many (x+d,2−d)(x + d, 2 – d) so that f((x+d,2−d))=x+d+2−d−2=xf((x+d, 2-d)) = x+d + 2-d -2 = x.

Thus for any g:R→R2g:\mathbb R \rightarrow R^2 so that f(a,b)=xf(a,b) = x and g(x)=(a,b)g(x) = (a,b) we would also have f(a+d,b−d)=a+d+b−d−2=a+b−2=f(a,b)=xf(a+d, b-d)= a+d +b-d -2 = a+b-2 = f(a,b) = x but g(f(a+d,b−d))=(a,b)≠(a+d,b−d)g(f(a+d,b-d)) = (a,b) \ne (a+d,b-d) so there can be no inverse function.

But that’s not an algebra question.

Thanks for the fast replies.

– Hector Isidor

Oct 20 at 19:38

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While this question is wrongly translated to English, it stated that the operation a*b = a+b-2 for a,b in R but * is in R2. This is the part that confuse me as well.

– Hector Isidor

Oct 20 at 19:41

If a,b∈R2a,b\in\mathbb R^2, then a+b−2a+b-2 doesn’t make any sense, because 2∉R22\not\in \mathbb R^2

– Thomas Andrews

Oct 20 at 20:03

The usual way of notating binary relations over UU to VV is ∗:U×U→V*:U \times U \to V, which can just as sensibly be written as ∗:U2→V*:U^2 \to V. One can say that ∗* is an operator over U2U^2, or that it is a binary operator over UU.

– Omnomnomnom

Oct 20 at 20:21

So…. we want to find e∈Re \in R so that (a,e)→a+e−2=a(a,e)\rightarrow a+e – 2 = a? That’d be 2. Not sure I see the point. The question is still wrong, isn’t it?

– fleablood

Oct 20 at 20:29

Very probably you intended the question to be about a binary operation ∗* on R\mathbb R, and “over \mathbb R^2\mathbb R^2” just refers to the fact that such an operation is a function with domain \mathbb R^2\mathbb R^2. Assuming that’s indeed your intention, the formula a*b=a+b-2a*b=a+b-2 makes sense; it defines a binary operation ** on \mathbb R\mathbb R. But you won’t be able to prove that it has no identity, because in fact 22 is an identity for this operation. Just compute: a*2=a+2-2=aa*2=a+2-2=a and 2*b=2+b-2=b2*b=2+b-2=b.

See the comment thread below the op….it seems that *: \mathbb R^2 \to R*: \mathbb R^2 \to R such that for every (a, b) \in \mathbb R^2(a, b) \in \mathbb R^2, a*b= a + b -2 \in \mathbb R.a*b= a + b -2 \in \mathbb R. I find it easier to simply note that we have f: \mathbb R^2\to \mathbb Rf: \mathbb R^2\to \mathbb R such that f(a, b) = a+b -2.f(a, b) = a+b -2.

– amWhy

Oct 20 at 20:34

But then what is the question? f(a,b) = a+b-2. What are we supposed to answer about it? Are we just supposed to note “Yes, that’s a function, all right”?

– fleablood

Oct 20 at 21:37

@fleablood I took the question to be “I have to prove that there is no identity element for this operation” (the OP’s second sentence). See also the title of the question.

– Andreas Blass

Oct 21 at 2:14

But… there IS an identity element for the operation! You can not prove there is not one when there is!

– fleablood

2 days ago

See… the problem is the OP got confused because he prove 2 was the identity. I asked what an operation “over R^2” was supposed to mean and suggested maybe the question was misinterpreted. Somewhere along the way if was noted that * = * = binary operation on R. Is the same as a*b = f(a,b); f:R\times R \rightarrow Ra*b = f(a,b); f:R\times R \rightarrow R. amWhy and the poor confused OP seemed to think that utterly changed the question to something else entirely. But it doesn’t. f transfers the question to the far less direct but equivalent: “Prove there is no e so that f(a,b)=a for all a”. That there IS is still an issue.

– fleablood

2 days ago

If we want to be very certain of what we are dealing with:

Define f:\mathbb{R}\to\mathbb{R}f:\mathbb{R}\to\mathbb{R} by f(x) = x-2f(x) = x-2. Then ff is a bijection, and f(a\ast b) = f(a) + f(b)f(a\ast b) = f(a) + f(b) for all a,b\in\mathbb{R}a,b\in\mathbb{R}.

This shows that the structure (\mathbb{R},\ast)(\mathbb{R},\ast) is equivalent to the structure (\mathbb{R},+)(\mathbb{R},+), and inherits all its properties. Since (\mathbb{R},+)(\mathbb{R},+) is an abelian group with identity 00, (\mathbb{R},\ast)(\mathbb{R},\ast) is an abelian group with identity f^{-1}(0) = 2f^{-1}(0) = 2.

1

Have y0u even followed the comment thread below the post; we are way past that, and you’ve misinterpreted what the OP clarified in a comment made almost thirty minutes ago, verified by OP 20 minutes ago. Your post doesn’t bear on the question intended.

– amWhy

Oct 20 at 20:47

@amWhy Yes, I read the comments and I understand your interpretation of the question. If you believe that you are right, your energy is probably best spent coming up with a clarifying edit to the question. However, in my view I am using the standard definition of an identity of a binary operation, which is usually defined to be a function S^2\to SS^2\to S for some set SS.

– Slade

Oct 21 at 2:59

@amWhy More specifically, the post did not specify whether the operation was arity 2 or arity 1. You say 1, I say 2. I don’t really see a way of resolving this without the input of the person who wrote the question.

– Slade

Oct 21 at 3:04